Expected number of noodles come from linearity of expectation. Probability comes from multiplying independent probabilities.

First calculate expected number of loops:
Denote X1 = identity variable, which takes value 1 every time a (long) noodle becomes a loop by connecting to its end, (if its own end is connected to some other noodle, then that other noodle's end will be considered). Thus, number of loops = X1 +...+ X100, thus, expected number of loops = E(X1) + ... + E(X100). This will give answer = 1/(2N-1) + 1/(2N-3) + ... + 1/3 + 1
where N = 100. This formula can also be proved using induction.

Calculating Probability of single large loop:
If you followed last result, this one is simple, this time all the identitiy variables are zero except last one. Thus probability is [1 - 1/(2N-1)]*[ 1 - 1/(2N-3)]* ... *[1-1/3]
= Product ( i = 2 to N) [ (2i-2) / (2i-1)]

Linearity of expectation

E(n) = n[1-(1-1/n)^n]

Notice that number of distinct points in the produced set is same as number of distinct points selected from the given set {1..n} . For that we denote indicator function Si, that is, Si = 1 if the interger i is taken into produced set.

Let S denote the number of distinct points selected, S = S1+..+Sn
E(S) = E(S1) +...+ E(Sn)
and E(Si) = 1*P(Si) = 1 - (Probability that 'i' is not chosen in any draw)

= 1 - ( Prob that is i is not chosen in one 1st draw)^n = 1 - ( 1 - 1/n )^n

Thus E(S) = n * ( 1 - ( (n-1) / n)^n )

Hint: the locations of the four aces in the deck divide it into the five groups X1, ...,X5.

53/5

Define X1,X2…X48, such that Xi = 1 if ith card turns over before 4 aces, 0 otherwise. Thus, total number of cards turned to see first ace = 1 + sum (Xi)
using linearity of expectation, E(X) = 1 + sum ( E(Xi))
Now consider the ith card and the four aces, all the orders are equally likely: X,A,A,A,A Or A,X,A,A,A, or 3 others. They are equally likely when we have no knowledge about position of these Aces with respect of the deck. Hence we have 5 equally likely slots for ith card: 1 A 2 A 3 A 4 A 5. We are interested in 1st slot.
Hence E(Xi) = 1/5 for all i = 1 to 48
Thus E(X) = 1 + 48/5 = 53/5 = 10.6

This can be done by recursive equation f(n) = (4/n)*1 + ((n-4)/n)*(1+f(n-1), where f(n)=expected cards to flip in a deck of n cards, but I definitely can't solve this without a computer.

A shorter explanation is to consider the 52 cards uniformly distributed over (0,1), so on average they're at k/53 for k=1,2,3,…,52. The four aces are on average at 1/5, 2/5, 3/5, 4/5. So 0.2=k/53 implies k=10.6, done!

Suppose x & y are outcomes of blackbox. [ x*cos(2 pi y) , x* cos(2 pi y) ] is not a uniform point on disc. Neither is the point (x,y)/sqrt(x^2+y^2) uniform.

theta = 2pi*x and r = squareroot(y). the take the point as (r*cos(theta),r*sin(theta))
Explanation:
The idea is that the probability of a point to be between r and r+dr is 2pi*r*dr. Hence to choose a radius we need to generate a number from random number generator which follows probability distribution 2*r.To generate random number which follows above probability distribution which follows uniform distribution we simply take the square root of the number generated by uniform distribution.